Final answer:
The escape speed from the surface of a planet with 1/6 the acceleration due to gravity of Earth and 1/3 its radius is approximately 3.73 km/s, calculated by adjusting the escape velocity formula using the given proportions relative to Earth.
Step-by-step explanation:
To calculate the escape speed from the surface of a planet where the acceleration due to gravity is 1/6th that of Earth and the planet's radius is 1/3rd of Earth's radius, we use the formula for escape velocity:
Escape velocity (v) = √(2*g*r),
where g is the acceleration due to gravity and r is the radius of the planet.
On Earth, the escape velocity is about 11.2 km/s. Since the acceleration due to gravity on the planet in question is 1/6 that of Earth and the radius is 1/3 that of Earth, we can substitute these values into the formula. Given that Earth's standard gravity is approximately 9.8 m/s², the planet's acceleration due to gravity would be 9.8/6 m/s², and its radius would be 1/3 of Earth's radius. When we substitute these values into the escape velocity formula, we get:
√(2*(9.8/6)*(Earth's radius/3))
This simplifies to √(2/18 * 9.8 * Earth's radius/3). Since Earth's escape velocity is 11.2 km/s with its full acceleration due to gravity and radius, we can simplify the formula further to:
√(1/9 * 11.2² km/s)
This gives us about 11.2 / 3 km/s or 3.73 km/s. Therefore, the escape speed from the planet in question is 3.73 kilometers per second.